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I have been given the following three non collinear-points $A(x_0,y_0,z_0)$, $B(x_1,y_1,z_1)$, $C(x_2,y_2,z_2)$.

I have also $\vec{AB}=<a,b,c>$, as well as $\vec{BC}=<d,e,f>$. The dot product between both vectors can be limited in the sense that $\vec{AB} \times \vec{BC}\ne 0 \therefore \mathrm{they \ are \ not \ parallel \ vectors.}$ Hence, once that is established since the vectors are not parallel, I achieved a cross product as from the following vectors, and got the following: \begin{align} \vec{n} &= \vec{AB}\ \times \ \vec{BC} \\ \vec{n}&= <g,h,i> \end{align}

Confusion part

I have this normal vector right, and then the next procedure would be to do the following: $r_0 = \mathrm{\ some \ point \ on \ the \ plane}$, and $r= <x,y,z>$.

The textbook gives this equation:

$$\vec{n}\ \cdot \ (r-r_0) = 0$$

Would my cross product be wrong if I did it like that or would the equation of the plane be wrong if I choose any point either A, B, or C?

Community
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Jose M Serra
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2 Answers2

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I 'm not sure what "any point either A,B, or C" means. Your cross-product is correct. For your point with position vector $r_0$ you can use $A$ or $B$ or $C$ or any other point on the plane.Your method is perfectly good but there are other good methods. For instance, you could write the equation of the plane as $$\alpha x+\beta y+\gamma z + \delta=0$$, substitute the coordinates of each of the known points and obtain a system of equations to be solved for $\alpha,\beta,\gamma,\delta.$

P. Lawrence
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Remember that there’s no such thing as “the” equation of a plane: there is an infinite number of them. If you end up with a linear equation, the only way that it could be wrong is if the three given points don’t all satisfy the equation.

Taking your second concern first, let’s rewrite the equation a bit: $$\vec n\cdot r = \vec n\cdot r_0.$$ What this equation says is that for all points $r$ on the plane, the dot product with $\vec n$ is constant. If you choose any other point on the plane as the fixed point $r_0$, then you will get exactly the same value for this dot product, and so also exactly the same equation. In short, it doesn’t matter which point you choose for $r_0$.

As to your first concern, recall that multiplying both sides of an equation by the same nonzero constant doesn’t change its solution set. So, both $\vec n\cdot(r-r_0)=0$ and $(c\vec n)\cdot(r-r_0)=0$ represent the same plane. In other words, the normal to a plane is not unique: you can use any vector that points in the right direction for $\vec n$ in the equation. Now, $\overrightarrow{BC}\times\overrightarrow{AB}=-\overrightarrow{AB}\times\overrightarrow{BC}$, so the order in which you multiply the two vectors doesn’t matter. Using $\overrightarrow{BA}$ instead of $\overrightarrow{AB}$ just changes the sign of the result as well, so the direction of the vectors between the points doesn’t matter, either. What if you choose a different pair of vectors, say, $\overrightarrow{AB}$ and $\overrightarrow{AC}$? Well, $$\overrightarrow{AB}\times\overrightarrow{BC} = (B-A)\times(C-B) = (B\times C)-(A\times C)+(A\times B)$$ and $$\overrightarrow{AB}\times\overrightarrow{AC} = (B-A)\times(C-A) = (B\times C)-(B\times A)-(A\times C)$$ but $A\times B=-(B\times A)$, therefore $\overrightarrow{AB}\times\overrightarrow{BC} = \overrightarrow{AB}\times\overrightarrow{AC}$. If you try out the other combinations, you’ll find that, up to a change in sign, you’ll always end up with the same normal. (This is related to the fact that changing the order of the arguments to the scalar triple product $A\times B\cdot C$ at most changes the sign of the result.) In short, it doesn’t matter which pairs of points you choose for the vectors in the cross product for $\vec n$. Now, if you were to use some other points on the plane, you might get a normal with a different length, but it will still point in the right direction, so the equation of the plane that you end up with will be equivalent.

There are, of course, other ways to formulate a Cartesian equation of a plane. For instance, given the coordinates of three noncolinear points, we can write the following equation directly: $$\begin{vmatrix}x&y&z&1\\x_1&y_1&z_1&1\\x_2&y_2&z_2&1\\x_3&y_3&z_3&1\end{vmatrix} = 0.$$ Swapping rows of a matrix only changes the sign of its determinant, so it’s obvious that in this form, the order of the points doesn’t matter. See this question and its answers if you’re curious as to why this determinant equation represents the plane through the three points.

amd
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